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Class 11 RD Sharma Solutions – Chapter 17 Combinations- Exercise 17.1 | Set 2

  • Last Updated : 11 Feb, 2021

Question 11. If 28C2r : 24C2r-4=225:11, find r.

Solution:

(28!/(28-2r)!2r!)/(24!/(24-2r+4)!(2r-4)!)=225/11

28x27x26x25/2r(2r-1)(2r-2)(2r-3)=225/11

28x27x26x25x11=225x(2r)(2r-1)(2r-2)(2r-3)

28x3x26x11=2r(2r-1)(2r-2)(2r-3)



14x2x3x13x2x11=2r(2r-1)(2r-2)(2r-3)

14x13x12x11=2r(2r-1)(2r-2)(2r-3)

2r=14

r=7

Question 12. If nC4,nC5 and nC6 are in AP, then find n.

Solution:

We know that the A.P series is represented as a,a+d,a+2d,..

=>2.nC5=nC4+nC6

=>2(n!/(n-5)!5!)=(n!/(n-4)!4!)+n!/(n-6)!6!



=>2/(n-5)5=(1/(n-5)(n-4))+(1/30)

=>(2/(n-5)5)-(1/(n-5)(n-4))=1/30

=>(2(n-4)-5)30=(n-5)(n-4)5

=>60n-240-150=(n-5)(n-4)5

=>12n-78=(n-5)(n-4)

=>12n-78=n2-9n+20

=>n2-21n+98=0

=>n2-7n-14n+98=0

=>n(n-7)-14(n-7)=0

=>(n-7)(n-14)=0

=>n=7, 14

Question 13. If 2nC3:nC2=44:3, find n.

Solution:

(2n!/(2n-3)!3!)/(n!/(n-2)!2!)=44/3

2n(2n-1)(2n-2)/3n(n-1)=44/3

2n(2n-1)(2n-2)=44n(n-1)

(2n-1)(2n-2)=22n-22

4n2-6n+2=22n-22

4n2-28n+24=0

n2-7n+6=0

(n-1)(n-6)=0

n=1 , 6

Let n=1

then 2(1)C3:2C2 is not possible because n<r.

So, n=6.

Question 14. If 16Cr=16Cr+2, find rC4.

Solution:

16=r+r+2

16=2r+2

14=2r

r=7

=>7C4=7!/3!4!



=5x6x7/3×2

=35

Question 15. If ∝=mC2, then find the value of C2.

Solution:

C2=∝!/(∝-2)!2!

=(∝-1)(∝)/2

=(mC2-1)(mC2)/2

=(m!/(m-1)!2!-1)(m!/(m-2)!2!)/2

=(m(m-1)-2)(m(m-1))/8

=(m2-m-2)(m(m-1))/8

=(m+1)(m-1)(m-2)m/8

=1/8[(m-2)(m-1)(m)(m+1)]

Question 16. Prove that the product of 2n consecutive negative integers is divisible by (2n)!.

Solution:

Let the 2n consecutive negative integers be -k,-k-1,-k-2,…,-k-(2n-1)

product of 2n consecutive negative integers=-kx-k-1x-k-2x….x-k-(2n-1)

=(-1)2nxkxk+1xk+2x…..xk+(2n-1)

=(-1)2nxkxk+1xk+2x…..xk+(2n-1)x(k-1)!/(k-1)!

=(-1)2nx(k+2n-1)!/(k-1)!

=(-1)2nx(k+2n-1)!(2n)!/(k-1)!(2n)!

=(-1)2nxk+2n-1C2nx(2n)!

Therefore the product of 2n consecutive negative integers is divided by (2n)!.

Question 17. For all positive integers n, show that 2nCn+2nCn-1=1/2[2n+2Cn+1].

Solution:

LHS=2nCn+2nCn-1

=2n!/n!(2n-n)!+2n!/(n-1)!(2n-n+1)!

=2n!/n!n!+2n!/(n-1)!(n+1)!

=2n!/n(n-1)!n!+2n!/(n-1)!n!(n+1)

=2n!/n(n+1)(n-1)!n![n+1+n]

=2n!(2n+1)/n(n+1)(n-1)!n!

=(2n+1)!/n!(n+1)!

=(2n+2)(2n+1)!/n!(n+1)!(2n+2)

=(2n+2)!/n!(n+1)!(n+1)2



=(2n+2)!/(n+1)!(n+1)!2

=(2n+2)!/(n+1)!(2n+2-n-1)!2

=1/2[2n+2Cn+1]

= RHS

Question 18. Prove that: 4nC2n:2nCn=[1x 3 x 5 x ….. x 4n-1]:[1 x 3 x 5 x …… x 2n-1]2.

Solution:

LHS=4nC2n/2nCn

=(4n!/2n!2n!)/(2n!/n!n!)

=[1 x 2 x ……x 4n] x[1 x2x3x….xn]2/[1x2x…x2n]3

=[1x3x5x…4n-1][2x4x6x…4n](n!)(n!)/[1x3x5x…x2n-1]2[2x4x6x…x2n]2(2n!)

=[1x3x5x..x4n-1]22n[1x2x3x..2n](n!)(n!)/[1x3x5x…x2n-1]2x22n[1x2x3x..xn]2(2n)!

=[1x3x5x…x4n-1]/[1x3x5x2n-1]2

=RHS

Question 19. Evaluate

_{5}^{20}\textrm{C} + \sum_{r=2}^{5}\ _{4}^{25-r}\textrm{C}

Solution:

=>20C5+20C4+21C4+22C4+23C4

W.K.T nCr+nCr-1=n+1Cr

=>21C5+21C4+22C4+23C4

=>22C5+22C4+23C4

=>23C5+23C4

=> 24C5



Question 20.1. Let r and n be positive integers such that 1<=r<=n. Then prove the following:

nCr/nCr-1=n-r+1/r

Solution:

LHS=(n!/r!(n-r)!)/(n!/(r-1)!(n-r+1)!

=(n-r)!(n-r+1)(r-1)!/(n-r)!(r)(r-1)!

=n-r+1/r

=RHS

Question 20.2. Let r an dn be positive integers such that 1<=r<=n. Then prove the following:

nxn-1Cr-1=(n-r+1)nCr-1

Solution:

LHS=n(n-1)!/(r-1)!(n-1-r+1)!

=n!/(r-1)!( n-r)!

=n!(n-r+1)/(r-1)!( n-r+1)(n-r)!

=n!(n-r+1)/(r-1)!(n-r+1)!

=(n-r+1)nCr-1

=RHS

Question 20.3. Let r and n be positive integers such that 1<=r<=n. Then prove the following:

nCr/n-1Cr-1=n/r

Solution:

LHS=(n!/r!(n-r)!)/(n-1)!/(n-r)!(r-1)!

=n(n-1)!(r-1)!(n-r)!/r(r-1)!(n-r)!(n-1)!

=n/r

=RHS

Question 20.4. Let r and n be positive integers such that 1<=r<=n. Then prove the following:

nCr+2 x nCr-1+ nCr-2=n+2Cr.

Solution:

W.K.T nCr+nCr-1=n+1Cr

LHS=nCr+nCr-1+nCr-1+nCr-2

=n+1Cr+n+1Cr-1

=n+2Cr

=RHS

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